Hdu 5302 Connect the Graph 2015 Multi-University Training Contest 2

Connect the Graph

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 108    Accepted Submission(s): 36
Special Judge

Problem Description

Once there was a special graph. This graph had n vertices and some edges. Each edge was either white or black. There was no edge connecting one vertex and the vertex itself. There was no two edges connecting the same pair of vertices. It is special because the each vertex is connected to at most two black edges and at most two white edges. 

One day, the demon broke this graph by copying all the vertices and in one copy of the graph, the demon only keeps all the black edges, and in the other copy of the graph, the demon keeps all the white edges. Now people only knows there are w0 vertices which are connected with no white edges, w1 vertices which are connected with 1 white edges, w2 vertices which are connected with 2 white edges, b0 vertices which are connected with no black edges, b1 vertices which are connected with 1 black edges and b2 vertices which are connected with 2 black edges.

The precious graph should be fixed to guide people, so some people started to fix it. If multiple initial states satisfy the restriction described above, print any of them.

 

Input

The first line of the input is a single integer T (T≤700), indicating the number of testcases. 

Each of the following T lines contains w0,w1,w2,b0,b1,b2. It is guaranteed that 1≤w0,w1,w2,b0,b1,b2≤2000 and b0+b1+b2=w0+w1+w2. 

It is also guaranteed that the sum of all the numbers in the input file is less than 300000.

 

Output

For each testcase, if there is no available solution, print −1. Otherwise, print m in the first line, indicating the total number of edges. Each of the next m lines contains three integers x,y,t, which means there is an edge colored t connecting vertices x and y. t=0 means this edge white, and t=1 means this edge is black. Please be aware that this graph has no self-loop and no multiple edges. Please make sure that 1≤x,y≤b0+b1+b2.

 

Sample Input

21 1 1 1 1 11 2 2 1 2 2

 

Sample Output

-161 5 04 5 02 4 01 4 11 3 12 3 1

 

Source

2015 Multi-University Training Contest 2

 

题目意思：

要求构造一组图的解，满足白色边：度为0的点个数为a[0],度为1的点个数为a[1]，度为2的点的个数为a[2]。

                                                                黑色边：度为0的点个数为b[0],度为1的点个数为b[1]，度为2的点的个数为b[2]。

输入a[0],a[1],a[2],b[0],b[1],b[2] 。且满足sigma(a[i]) = sigma(b[i])

思路：总共点数为sigma(a[i])。若a[1]&1 或b[1]&1 则无解。度为1的点必须成对存在，在这只有0 1 2度数的图中。

清楚白色边和黑色边不相互影响，他们是独立的。

总的边数为总度数/2 。即：（a[1]+a[2]*2）/2.

先构造白边度数为2的点，必然是（1,2）（2,3）（3,4）....(a[2]+1,a[2]+2)..如此下去，直到完成度数2的节点的构造。

再够着白边度数为1的点，(a[2]+3,a[2]+4) (a[2]+5,a[2]+6).......

最后白色边度数为0的点，（,）(,)...(a[0]+a[1]+a[2],a[0]+a[1]+a[2])

黑色边的处理需要间隔2的跳变（图中两点之间只有一条边），其余和白色边构造类似。

#include<iostream>#include<cstdio>#include<cmath>#include<cstring>using namespace std;#define sf scanfint T;int a[5],b[5];int d[1000005];int main(){    //freopen("c.in","r",stdin);    sf("%d",&T);    while(T--)    {        for(int i = 0;i<3;i++)sf("%d",a+i);        for(int i = 0;i<3;i++)sf("%d",b+i);        int sum = 0;        for(int i = 0;i<3;i++)sum+=a[i];        if( (a[1]&1) || (b[1]&1) )        {            puts("-1");            continue;        }        if(sum==4)        {            puts("4\n1 2 0\n1 3 0\n2 3 1\n3 4 1");            continue;        }        printf("%d\n",a[1]/2+a[2]+b[1]/2+b[2]);        int t = 1;        while(a[2]!=-1){printf("%d %d 0\n",t,t+1);t++;a[2]--;}t++;        while(a[1]!=2){printf("%d %d 0\n",t,t+1);t+=2;a[1]-=2;}            int tt = 0;            for(int i = 1;i<=sum;i+=2) d[tt++] = i;            for(int i = 2;i<=sum;i+=2)d[tt++] = i;        t = 0;        while(b[2]!=-1){printf("%d %d 1\n",min(d[t],d[t+1]),max(d[t],d[t+1]));t++;b[2]--;}t++;        while(b[1]!=2){printf("%d %d 1\n",min(d[t],d[t+1]),max(d[t],d[t+1]));t+=2;b[1]-=2;}    }}